During the thermodynamic process shown in the | vedclass

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(D) The correct option is $(d)$.For a straight $P-V$ graph line passing through the origin,$P \propto V$,which means $P = kV$ for some constant $k > 0

Vedclass · Accepted Answer

$\Delta U > 0$

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(D) The correct option is $(d)$.For a straight $P-V$ graph line passing through the origin,$P \propto V$,which means $P = kV$ for some constant $k > 0$.From the ideal gas equation,$PV = nRT$,we can substitute $P = kV$ to get $kV^2 = nRT$,or $T = \frac{k}{nR} V^2$.As the volume $V$ increases along the arrow,the temperature $T$ also increases,so $\Delta T > 0$. Thus,$\Delta T 
eq 0$.Since the volume is increasing,the work done by the gas is positive,$W = \int P dV > 0$.Since the temperature increases,the internal energy of the ideal gas increases,$\Delta U = nC_v \Delta T > 0$.Therefore,$\Delta U > 0$ is the correct statement.

During the thermodynamic process shown in the figure for an ideal gas:

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